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2021 AMC 10B Problems/Problem 18

Revision as of 15:59, 11 July 2024 by Goldroman (talk | contribs) (→‎Solution 3)

Problem

A fair $6$-sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?

$\textbf{(A)} ~\frac{1}{120} \qquad\textbf{(B)} ~\frac{1}{32} \qquad\textbf{(C)} ~\frac{1}{20} \qquad\textbf{(D)} ~\frac{3}{20} \qquad\textbf{(E)} ~\frac{1}{6}$

Solution 1

Since 3 out of 6 of the numbers are even, there is a $\frac36$ chance that the first number we choose is even.

Since the number rolled first is irrelevant, we don't have to consider it. Therefore there are 2 even numbers out of the 5 choices left. There is a $\frac{2}{5}$ chance that the next number that is distinct from the first is even.

There is a $\frac{1}{4}$ chance that the next number distinct from the first two is even. (There is only one even integer left. ) With all the even integers taken, the next integer rolled must be odd.

$\frac{3}{6} \cdot \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{20}$, so the answer is $\boxed{\textbf{(C) }\frac{1}{20}}.$

~Tucker

Solution 2

Every set of three numbers chosen from $\{1,2,3,4,5,6\}$ has an equal chance of being the first 3 distinct numbers rolled.

Therefore, the probability that the first 3 distinct numbers are $\{2,4,6\}$ is $\frac{1}{{6 \choose 3}}=\boxed{(C)~\frac{1}{20}}$

~kingofpineapplz

Solution 3

Note that the problem is basically asking us to find the probability that in some permutation of $1,2,3,4,5,6$ that we get the three even numbers in the first three spots.

There are $6!$ ways to order the $6$ numbers and $3!(3!)$ ways to order the evens in the first three spots and the odds in the next three spots.

Therefore the probability is $\frac{3!(3!)}{6!} = \frac{1}{20} = \boxed{\textbf{(C)}}$. ~abhinavg0627

Solution 4

Let $P_n$ denote the probability that the first odd number appears on roll $n$ and all our conditions are met. We now proceed with complementary counting.

For $n \le 3$, it's impossible to have all $3$ evens appear before an odd. Note that for $n \ge 4,$ \[P_n = \frac {1}{2^{n}} - \frac {1}{2^{n}} \left(\frac{\binom{3}{2}(2^{n-1}-2)+\binom{3}{1}}{3^{n-1}}\right)\] since there's a $\frac {1}{2^{n}}$ chance that the first odd appears on roll $n$ (disregarding the other conditions) and the other term is subtracting the probability that less than $3$ of the evens show up before the first odd roll. Simplifying, we arrive at \[P_n= \frac {1}{2^{n}} - \left(\frac {3(2^{n-1})-3}{2^{n} \cdot 3^{n-1}}\right) = \frac {1}{2^{n}} - \frac {1}{2 \cdot 3^{n-2}} + \frac{1}{2^{n} \cdot 3^{n-2}}.\]

Summing for all $n$, we get our answer of \[\left (\frac {1}{2^{4}} + \frac {1}{2^{5}} + ... \right) - \left (\frac {1}{2 \cdot 3^{2}} + \frac {1}{2 \cdot 3^{3}} + ... \right) + \left (\frac {1}{2^{4} \cdot 3^{2}} + \frac {1}{2^{5} \cdot 3^{3}} + ... \right) = \left (\frac {1}{8} \right) - \left(\frac {\frac {1}{18}}{ \frac{2}{3}} \right) + \left(\frac {\frac {1}{144}}{\frac {5}{6}} \right) = \frac {1}{8} - \frac {1}{12} + \frac{1}{120} = \boxed{\textbf{(C) }\frac{1}{20}.}\]

~ike.chen

Solution 5 (States)

Let $E_n$ be that probability that the condition in the problem is satisfied given that we need $n$ more distinct even numbers. Then, \[E_1=\frac{1}{6}+\frac{1}{3}\cdot E_1+\frac{1}{2}\cdot 0,\] since there is a $\frac{1}{3}$ probability that we will roll an even number we already have rolled and will be in the same position again. Solving, we find that $E_1=\frac{1}{4}$.

We can apply the same concept for $E_2$ and $E_3$. We find that \[E_2=\frac{1}{3}\cdot E_1+\frac{1}{6}\cdot E_2+\frac{1}{2}\cdot 0,\] and so $E_2=\frac{1}{10}$. Also, \[E_3=\frac{1}{2}\cdot E_2+\frac{1}{2}\cdot 0,\] so $E_3=\frac{1}{20}$. Since the problem is asking for $E_3$, our answer is $\boxed{\textbf{(C) }\frac{1}{20}}$. -BorealBear

Solution 6 (Infinite Geometric Sequence Method)

Let's say that our even integers are found in the first $n$ numbers where n must be greater than or equal to $3$. Then, we can form an argument based on this. There are $3^n$ total ways to assign even numbers being $($2$,$4$,$6$)$ to each space. Furthermore, we must subtract the cases where we have only a single even integer present in the total $n$ spaces and the case where there are only $2$ distinct even integers present. There is $1$ way we can have a single even integer in the entire $n$ spaces, therefore giving us just $1$ option for each of the three integers, so we have $3$ total cases for this. Moreover, the amount of cases with just $2$ distinct even integers is $2^n$ but subtract the cases where all of the n spaces is either a single integer giving us $2^n-2$, but we multiply by $3C2$ because of the ways to choose $2$ distinct even integers that are used in the sequence of $n$. Finally, we have $\sum_{n=3}^{\infty} \frac{(3^n-3-3(2^n-2))}{6^n}$ note: we must divide all of this by $6^n$ for probability. Additionally, over the entire summation, we multiply by $1/2$ because of the $1/2$ probability of selecting an odd at the end of all the evens. Therefore, if you compute this using infinite geometric sequences, you get $1/20$. $\boxed{\textbf{(C) }\frac{1}{20}.}$

~Jske25

Solution 7 (Complementary + PIE)

We know that the sequence with odd ending will eventually add up to probability 1.

For every even number appears at least once, let's look at the complementary, i.e. without some even numbers ${2, 4, 6}$. $A = \{ \text{without} 2\}, B = \{ \text{without} 4\}, C = \{ \text{without} 6\}$.

$1 - A\cup B\cup C = 1 - (A + B + C - A\cap B - B\cap C - A\cap C + A\cap B\cap C)$

Let $f(x) = \frac{1}{2} + x\frac{1}{2} + x^2\frac{1}{2} + \ldots = \frac{1}{2} \frac{1}{1-x}$

2 even numbers left $|A|=|B|=|C|: f(\frac{2}{6}) = \frac{1}{2} \frac{1}{1-\frac{1}{3}} = \frac{3}{4}$,

1 even number left $|A\cap B| = |B\cap C| = |A\cap C|: f(\frac{1}{6}) = \frac{1}{2} \frac{1}{1-\frac{1}{6}} = \frac{3}{5}$

0 even number left $|A\cap B\cap C|: f(0) = \frac{1}{2}$

ans $= 1 - A\cup B\cup C = 1- (\frac{3}{4} * 3 - \frac{3}{5} * 3 + \frac{1}{2}) = \frac{1}{20}$

~aliciawu

Solution 8

We don't care what happens after we pick the first three distinct even numbers. This can be done is $3! = 6$ ways. Totally there are $6 \cdot 5 \cdot 4$ ways to pick three distinct numbers. So the answer is $\frac{6}{6\cdot5\cdot4} = \frac{1}{20}$.

~coolmath_2018

Solution 9

We can use complementary counting for this problem by eliminating the chances that we get 0,1, and 2 even numbers before the odd number. To get 0 even numbers before the odd number has a probability of 3/6 or 1/2. To get 1 even number before the odd number has a probability of 3/6*3/5 which is 3/10 (0.3). To get the probability of 2 even numbers before the odd number it is 3/6*2/5*3/4 which is 3/20 (0.15). Now we eliminate all of these cases to get the answer to the question by doing 1-(0.5+0.3+0.15) which is equal to 0.05 or 1/20 so the answer is C 1/20

~HYPEBEAST1

Video Solution (🚀Under 1 min 🚀)

https://youtu.be/8amWKsbcjXs

~Education, the Study of Everything

Video Solution by OmegaLearn

https://youtu.be/IX-Y38KPxqs

~pi_is_3.14

Video Solution by hurdler (complementary probability)

https://www.youtube.com/watch?v=k2Jy4ni9tK8

Video Solution by TheBeautyofMath

https://youtu.be/FV9AnyERgJQ?t=480

~IceMatrix

Video Solution by Interstigation (Simple Bash With PIE)

(which stands for Principle of Inclusion and Exclusion) https://youtu.be/2SGmSYZ5bqU

~Interstigation

See Also

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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